📚 Maclaurin Series Expansion: AQA A-Level Maths Essentials | 麦克劳林展开 AQA A-Level 数学考点精讲
The Maclaurin series is a powerful tool in AQA A-Level Mathematics, allowing you to express complicated functions as infinite polynomials. It appears regularly in Pure Paper 2 and Paper 3, testing your ability to derive, manipulate, and apply standard expansions. Mastering this topic will boost your confidence in series work, limits, approximations, and even elementary differential equations.
麦克劳林级数是 AQA A-Level 数学中的重要工具，它能把复杂的函数表示成无穷多项式。这部分内容常在纯数学卷 2 和卷 3 中出现，考察你对标准展开式的推导、变换和应用能力。掌握好这一考点，能让你更从容地应对级数、极限、近似计算甚至简单微分方程的题目。
1. What is a Maclaurin Series? | 什么是麦克劳林级数？
The Maclaurin series of a function f(x) is its Taylor series expansion centred at 0. Provided f is infinitely differentiable near 0, the series is f(x) = f(0) + f'(0)x + f”(0)/2! x² + f”'(0)/3! x³ + … + f⁽ⁿ⁾(0)/n! xⁿ + … . In AQA papers, you are expected to find the first few non-zero terms by repeated differentiation.
函数 f(x) 的麦克劳林级数是它在 x=0 处的泰勒展开。只要 f 在 0 附近无限可导，级数就是 f(x) = f(0) + f'(0)x + f”(0)/2! x² + f”'(0)/3! x³ + … + f⁽ⁿ⁾(0)/n! xⁿ + … 。在 AQA 考题中，你需要通过反复求导来求出前几个非零项。
2. Step-by-Step Derivation Method | 逐步求导推导法
To build a Maclaurin series, first evaluate f(0) to f”'(0) or as required. For f(x) = eˣ, you get f(0)=1, f'(0)=1, f”(0)=1, so the series becomes 1 + x + x²/2! + x³/3! + … . For sin x, f(0)=0, f'(0)=1, f”(0)=0, f”'(0)=-1, giving x – x³/3! + x⁵/5! – … . Always check for cyclic patterns in derivatives.
构建麦克劳林级数时，先求出 f(0) 到 f”'(0) 或题目要求的阶数。例如 f(x)=eˣ，得到 f(0)=1, f'(0)=1, f”(0)=1，于是级数为 1 + x + x²/2! + x³/3! + … 。对 sin x，f(0)=0, f'(0)=1, f”(0)=0, f”'(0)=-1，得到 x – x³/3! + x⁵/5! – … 。解题时要注意导数是否存在循环规律。
3. Standard Expansions You Must Memorise | 必须牢记的标准展开式
AQA expects you to know these by heart and state their range of validity:
AQA 要求你熟记下列展开式并明确其有效范围：
| f(x) | Maclaurin series | Valid for |
| eˣ | 1 + x + x²/2! + x³/3! + … + xⁿ/n! + … | all real x |
| sin x | x − x³/3! + x⁵/5! − … + (−1)ⁿ x²ⁿ⁺¹/(2n+1)! + … | all real x |
| cos x | 1 − x²/2! + x⁴/4! − … + (−1)ⁿ x²ⁿ/(2n)! + … | all real x |
| ln(1+x) | x − x²/2 + x³/3 − … + (−1)ⁿ⁺¹ xⁿ/n + … | −1 < x ≤ 1 |
| (1+x)ⁿ (binomial) | 1 + nx + n(n−1)/2! x² + … + n(n−1)…(n−r+1)/r! xʳ + … | |x| < 1 (unless n is a positive integer) |
4. Validity and Convergence | 有效性与收敛域
Every series has a domain where it converges to the function. For eˣ, sin x and cos x the series converges for all real x. For ln(1+x) the expansion is valid only for −1 < x ≤ 1. For (1+x)ⁿ with non‑integer n, the series converges when |x| < 1. Always state the range of x when answering an exam question.
每个级数都有其收敛到原函数的定义域。eˣ、sin x 和 cos x 的级数对全体实数都收敛。ln(1+x) 的展开式仅在 −1 < x ≤ 1 内有效。当 (1+x)ⁿ 的指数 n 不是正整数时，级数在 |x| < 1 时收敛。答题时务必说明 x 的取值范围。
5. Substitution to Find New Expansions | 通过代换求新的展开式
You can obtain a new series by replacing x with a multiple or power. For example, e²ˣ = 1 + 2x + (2x)²/2! + (2x)³/3! + … , valid for all x. sin(3x) = 3x − (3x)³/3! + … . Be careful with brackets: 1/(1+2x) = 1 − 2x + 4x² − 8x³ + … for |2x| < 1, i.e. |x| < 1/2.
你可以通过把 x 替换成倍数或幂来获得新的级数。例如 e²ˣ = 1 + 2x + (2x)²/2! + (2x)³/3! + … ，对全体实数有效。sin(3x) = 3x − (3x)³/3! + … 。注意括号的使用：1/(1+2x) = 1 − 2x + 4x² − 8x³ + … 当 |2x| < 1，即 |x| < 1/2。
6. Multiplication and Division of Series | 级数的乘法与除法
Sometimes you need the product of two series, e.g. eˣ sin x up to x³. Expand each individually: eˣ ≈ 1 + x + x²/2 + x³/6, sin x ≈ x − x³/6. Multiply term by term and keep only terms up to x³: (1)·(x) + (x)·(x) + (1)·(−x³/6) + … giving x + x² + (1/2 − 1/6)x³ = x + x² + x³/3. Division can be done by comparing coefficients if the denominator series has a constant term.
有时你需要求两个级数的乘积，比如 eˣ sin x 展开到 x³。先分别展开：eˣ ≈ 1 + x + x²/2 + x³/6，sin x ≈ x − x³/6。逐项相乘并只保留到 x³ 次项：(1)·(x) + (x)·(x) + (1)·(−x³/6) + … 得到 x + x² + (1/2 − 1/6)x³ = x + x² + x³/3。如果分母级数含有常数项，可以用比较系数的方法进行除法。
7. Maclaurin Series and Limits | 麦克劳林级数与求极限
When evaluating an indeterminate form like (sin x − x)/x³ as x→0, expand sin x = x − x³/6 + … . Then (sin x − x)/x³ = (−x³/6 + …)/x³ = −1/6 + higher terms → −1/6. This technique is especially useful when L’Hôpital’s rule would be messy.
在计算类似 lim x→0 (sin x − x)/x³ 的不定式时，将 sin x 展开为 x − x³/6 + … ，则 (sin x − x)/x³ = (−x³/6 + …)/x³ = −1/6 + 高阶项 → −1/6。当洛必达法则计算繁琐时，这种方法尤其好用。
8. Approximations and Error Bounds | 近似值与误差估计
Truncating the series after a few terms gives a polynomial approximation. For small x, sin x ≈ x − x³/6 gives excellent accuracy. AQA may ask you to bound the error using the next term or Lagrange remainder. The alternating series error bound is very handy: the error is less than the first omitted term when the series alternates and terms decrease in magnitude.
截取级数的前几项就得到多项式近似。对于很小的 x，sin x ≈ x − x³/6 精度很高。AQA 可能要求用下一项或拉格朗日余项来估计误差。交错级数的误差估计非常实用：当级数正负交替且项单调递减时，误差不超过第一个被省略的项。
9. Term-by-Term Differentiation and Integration | 逐项微分与逐项积分
Knowing that sin x = x − x³/3! + x⁵/5! − …, you can differentiate term by term to obtain cos x = 1 − x²/2! + x⁴/4! − … . Similarly, integrating the geometric series 1/(1+x) = 1 − x + x² − x³ + … gives ln(1+x) = x − x²/2 + x³/3 − … (plus constant, which is 0). This is a valid approach within the interval of convergence.
已知 sin x = x − x³/3! + x⁵/5! − … ，你可以逐项微分得到 cos x = 1 − x²/2! + x⁴/4! − … 。类似地，对几何级数 1/(1+x) = 1 − x + x² − x³ + … 逐项积分就得到 ln(1+x) = x − x²/2 + x³/3 − … （加上常数，此处常数为 0）。在收敛区间内，这是完全合理的方法。
10. Exam Tips and Common Pitfalls | 考试技巧与常见错误
Many candidates lose marks by forgetting to divide by factorials, miswriting the signs, or omitting the range of validity. Always write the general term if asked. Do not confuse the expansion of ln(1+x) with that of ln(1−x). When using substitution, adjust the validity condition accordingly. Practise timed questions that ask for a series up to and including x⁴ to improve speed and accuracy.
很多考生因忘记除以阶乘、写错符号或漏写收敛域而丢分。如果题目要求，务必写出通项公式。不要把 ln(1+x) 和 ln(1−x) 的展开式混淆。使用代换时，要相应调整有效条件。多练习要求展开到 x⁴ 项的限时题，可以有效提升解题速度和准确性。
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